Question

Express: $Z=\frac{i-1}{\cos \frac{\pi }{3}+i\sin \frac{\pi }{3}}$ in polar form.

Answer 1

$Z=\frac{i-1}{cos\frac{\pi }{3}+isin\frac{\pi }{3}}$

$=\frac{i-1}{(\frac{1}{2}+i\frac{\sqrt{3}}{2})}\times \frac{(\frac{1}{2}-\frac{i\sqrt{3}}{2})}{(\frac{1}{2}-\frac{\sqrt{3}i}{2}}$

$=\frac{\frac{i}{2}+\frac{\sqrt{3}}{2}-\frac{1}{2}+\frac{\sqrt{3}i}{2}}{\frac{1}{4}+\frac{3}{4}}$

$=(\frac{\sqrt{3}}{2}-\frac{1}{2})+i(\frac{1}{2}+\frac{\sqrt{3}}{2})$

polor from

$z=r(cos\theta +isin\theta )$

$rcos\theta =\frac{\sqrt{3}-1}{2},rsin\theta =\frac{\sqrt{3}+1}{2}.$

squaring and adding

$\therefore r^2=\frac{3-2\sqrt{3}+1+3+2\sqrt{3}+1}{4}=2$

$\therefore r=\sqrt{2}$

$cos\theta =\frac{\sqrt{3}-1}{2\sqrt{2}}=\frac{\sqrt{3}}{2}\times \frac{1}{\sqrt{2}}-\frac{1}{2}\times \frac{1}{\sqrt{2}}$

$=cos\frac{\pi }{6}.cos\left(\frac{\pi }{4}\right)-sin\frac{\pi }{6}.sin\frac{\pi }{4}$

$cos\theta =cos(\frac{\pi }{6}+\frac{\pi }{4})$

$\therefore \theta =\frac{5\pi }{12}\left(cos\right(A+B\left)\right)=cosA.cosB-sinAsinB)$

Hence, required poloar from

$z=\sqrt{2}(cos\frac{5\pi }{12}+isin\frac{5\pi }{12})$